Mathematics - Functions Question with Solution | TestHub

$y=\frac{x^2-3x+4}{x^2+3x+4}$

$$y{x}^2+3xy+4y=x^2-3x+4$$

$$x^2(y-1)+3x(y+1)+4(y-1)=0$$

$$D\ge 0\Rightarrow 9(y+1{)}^2-4\times 4(y-1{)}^2\ge 0$$

$$(3(y+1)-4(y-1))(3(y+1)+4(y-1))\ge 0$$

$$(-y+7)(7y-1)\ge 0$$

$$(y-7)\left(y-\frac{1}{7}\right)\le 0$$

$$\frac{1}{7}\le y\le 7$$